jhenezhubby01ff

2022-09-05

According to my book the spring constant is given by $k={V}^{″}\left({X}_{o}\right)$. Where $V\left(X\right)$ is the potential energy function. if I use the function $V\left(X\right)={X}^{6}+{X}^{4}$ the spring constant is zero at ${X}_{o}=0$. However, the plot looks like a stable equilibrium that would allow harmonic oscillation near ${X}_{o}$. So why is k zero in this case?

Krha77

The force is given by the first derivative of the potential:
$F\left(X\right)=-\frac{d}{dX}V\left(X\right)=-6{X}^{5}-4{X}^{3}.$
At small displacements, i.e. when $X\ll 1$ we have ${X}^{5}\ll {X}^{3}$, that is the returning force is proportional to ${X}^{3}$ and not to X, as is in the case of a harmonic oscillator. The spring constant is indeed zero.
You are right that there are oscillations in this case, however these oscillations are not harmonic. Your logic based on the shape of the potential energy is a sound one and is used for analyzing nonlinear oscillators, such as the one that you have here.

samuelaplc